A111319 -id:A111319 - cp's OEIS frontend

A112271 One fifth of the sum of the first n primes, when an integer.

1, 2, 20, 32, 88, 212, 296, 344, 1070, 1166, 1374, 1655, 2248, 2698, 3368, 3730, 3916, 4936, 5160, 5388, 6725, 6983, 8788, 11338, 12382, 12923, 13480, 15026, 16244, 17717, 19033, 19481, 19937, 21108, 24584, 29191, 30345, 33008, 33921, 34850

Offset: 1

Jonathan Vos Post, Nov 30 2005

			a(1) = 1 = (2+3)/5 = A007504(2)/5 = 5/5.
a(2) = 2 = (2+3+5)/5 = A007504(3)/5 = 10/5.
a(3) = 20 = (2+3+5+7+11+13+17+19+23)/5 = A007504(9)/5 = 100/5.
a(4) = 32 = (2+3+5+7+11+13+17+19+23+29+31)/5 = A007504(11)/5 = 160/5.
a(5) = 88 = A007504(17)/5 = 440/5.
a(6) = 212 = A007504(25)/5 = 1060/5.
a(7) = 296 = A007504(29)/5 = 1480/5.
a(8) = 344 = A007504(31)/5 = 1720/5.

Bach, E. and Shallit, J. Sect. 2.7 in Algorithmic Number Theory, Vol. 1: Efficient Algorithms. Cambridge, MA: MIT Press, 1996.
H. L. Nelson, "Prime Sums", J. Rec. Math., 14 (1981), 205-206.

Harvey P. Dale, Table of n, a(n) for n = 1..1000
Leo Moser, Notes on number theory. III. On the sum of consecutive primes, Canad. Math. Bull. 6 (1963), pp. 159-161.
Eric Weisstein's World of Mathematics, Prime Sums.

Cf. A000040, A007504, A111319, A112040.

s = 0; lst = {}; Do[s = s + Prime[n]; If[Mod[s, 5] == 0, AppendTo[lst, s/5]], {n, 250}]; lst (* Robert G. Wilson v, Dec 04 2005 *)
Select[Accumulate[Prime[Range[400]]]/5,IntegerQ] (* Harvey P. Dale, May 03 2017 *)

{a(n)} = {A007504(k)/5 iff 5 | A007504(k)}. {a(n)} = {(p_1 + p_2 + ... + p_k)/5 iff the sum is an integer}. It is sufficient that A007504(k) == 0 (mod 10), but not necessary (the last five consecutive primes ending in 1 can give a solution). It is necessary that k = 2 or k is odd.

More terms from Stefan Steinerberger and Robert G. Wilson v, Dec 04 2005

cp's OEIS Frontend

A112271 One fifth of the sum of the first n primes, when an integer.

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